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Word Break/Word Segment

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

Dynamic programming。用一个table记录下0~i-1是否符合要求。注意一开始的table[0]=true,以及table长度为输入字符串长度加一。Loop 0~s.length()-1,只能从table[i]=true的地方开始,因为那之前都是匹配的才能接着匹配。

public class Solution {    public boolean wordBreak(String s, Set<String> dict) {        int strlen=s.length();        boolean [] table=new boolean[strlen+1];        table[0]=true;//begin is always true, so we can start the loop        for(int i=0;i<strlen;i++){            if(!table[i])// 0~i-1 are not words                continue;            for(String word: dict){                int len=word.length();                int end=i+len-1;                if(end>strlen-1)                    continue;                if(table[end+1])//already match at this position                    continue;                if(s.substring(i,end+1).equals(word)){                    table[end+1]=true;                }            }        }        return table[strlen];    }}